---
title: "Series sum using Z-transform"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:series-sum-using-z-transform"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/series-sum-using-z-transform
site: BCT Engineering
---

# Series sum using Z-transform

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2082 Kartik (new course), 2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/series-sum-using-z-transform
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Apply Z-transform to find the sum of the infinite series Σ_{k=0}^∞ e^(ak) for a > 0.

## Short answer

Z-transform of e^{ak} is z/(z-e^a). Setting z=1 yields Σ e^{ak}=1/(1-e^a). But for a>0 the series diverges; result valid only when a<0.

## Answer

The Z-transform of the sequence x(k) = e^(ak) is

```text
X(z) = Σ_{k=0}^∞ e^(ak) z^(−k) = Σ (e^a/z)ᵏ = z/(z − e^a),   |z| > e^a
```

The required series is X(z) evaluated at z = 1:

```text
Σ e^(ak) = X(1) = 1/(1 − e^a)
```

**Remark.** This is a geometric series with ratio r = e^a. For a > 0 we have r > 1, so the series actually diverges. The value 1/(1 − e^a) = −1/(e^a − 1) is only a formal result (z = 1 lies outside the region |z| > e^a). The series converges, with sum 1/(1 − e^a), when a < 0 (and for the usual form e^(−ak) with a > 0 the sum is 1/(1 − e^(−a)) = e^a/(e^a − 1)).

**Answer: formally 1/(1 − e^a); convergent only if a < 0.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Z[e^(ak)] =
   A. z/(z + e^a)
   B. z/(z - e^(-a))
   C. 1/(z - e^a)
   D. z/(z - e^a) for |z| > e^a
   Answer: D. Geometric series in e^a/z. (Key point: Z-transform)

2. The series sum is X(z) at
   A. z = 1
   B. z = 0
   C. z = e^a
   D. z to infinity
   Answer: A. z^(-k) = 1 at z = 1. (Key point: Evaluate)

3. For a > 0, why does the series diverge?
   A. The ratio is negative
   B. The ratio e^a > 1
   C. Because z = 0
   D. Because a is an integer
   Answer: B. A geometric series converges only if |r| < 1. (Key point: Convergence)

## Related in Mathematics III

- [Solve ODE by Laplace transform y''+2y'+5y=e^(−t)sint](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-ode-by-laplace-transform-y-six) — Mathematics III, chapter 3, asked 2×
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